Problem: On the first day of spring, an entire field of flowering trees blossoms. The population of locusts consuming these flowers rapidly increases as the trees blossom. The locust population gains $87\%$ of its size every $2.4$ days, and can be modeled by a function, $L$, which depends on the amount of time, $t$ (in days). Before the first day of spring, there were $1100$ locusts in the population. Write a function that models the locust population $t$ days since the first day of spring. $L(t) = $
Solution: The strategy We can model the situation with an exponential function of the general form A ⋅ B f ( t ) A\cdot B\^{ f(t)}, where $A$ is the initial quantity, $B$ is a factor by which the quantity is multiplied over constant time intervals, and $f(t)$ is an expression in terms of $t$ that determines those time intervals. Let's use the given information to determine $A$, $B$, and $f(t)$. Understanding what's given We are given that the initial number of locusts is $1100$, and the locust population gains $87\%$ of its size every $2.4$ days. Note that gaining $87\%$ is the same as being multiplied by $1.87$. [Why?] This means that the initial quantity is $A=1100$ and the factor is $B=1.87$. We need to find $f(t)$ based on the fact that the quantity is multiplied by $1.87$ every $2.4$ days. Finding the expression in the exponent We know that the locust population is multiplied by $1.87$ every $2.4$ days. This means that each time $t$ increases by $2.4$, $f(t)$ increases by $1$. Therefore, $f(t)$ is a linear function whose slope is $\dfrac{1}{2.4}$. When the initial measurement is made, the population hasn't changed. So $L(0) = 1100$, which means that $f(0)=0$. [Why?] Therefore, $f(t)$ must be $\dfrac{t}{2.4}$. Summary We found that the following function models the locust population $t$ days since the first day of spring. L ( t ) = 1100 ⋅ ( 1.87 ) t 2.4 L(t)=1100\cdot (1.87)\^{ \frac{t}{2.4}}